Let $E$ be a vector space on topological field $K$. Denote by $E^{*}$ the set of all $K$-linear maps of $E$ into $K$. BY the $K$-linear topology of $E$ is meant the least topology of $E$ with respect to which every element of $E^*$ is continuous.
Hence, if $W$ is an open neighborhood of $0\in K$, then $U=\{x\in E: y^*(x)\in W\}$ is an open neighborhood of $0\in E$ for every $y^*\in E^*$. This implies that if $E$ is finite dimensional vector space on $K$ with $K$-linear topology and $\{e_i:i=1, 2, \ldots, n\}$ be a base for $E$ then $U=\{x\in E: e^*_i(x)\in W, (i=1, 2, \ldots, n)\} $ is open neighborhood of $0\in E$.
What can say about the converse of it?
Is it true that if $E$ is finite dimensional vector space on $K$ and $U$ is an open neighborhood of $0\in E$, then there is open neighborhood $W$ of $0\in K$ such that $$U=\{x\in E: \langle x, e^*_i\rangle \in W, (i=1, 2, \ldots, n) \}$$